A cannon ball is fired with a velocity 200 , | vedclass

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(B) Let the mass of the cannon ball be $m$. At the highest point,the vertical component of velocity is $0$,and the horizontal component is $u_x = u \c

Vedclass · Accepted Answer

$300 \, m/s$ in the horizontal direction

Vedclass · Answer

(B) Let the mass of the cannon ball be $m$. At the highest point,the vertical component of velocity is $0$,and the horizontal component is $u_x = u \cos 	heta = 200 \cos 60^{\circ} = 200 	imes 0.5 = 100 \, m/s$.Thus,the momentum of the cannon ball at the highest point is $\vec{P} = m 	imes 100 \hat{i}$.After the explosion,the mass splits into $3$ equal fragments of mass $m/3$ each.Let the velocity of the third fragment be $\vec{V} = V_x \hat{i} + V_y \hat{j}$.The velocities of the fragments are $\vec{v}_1 = 100 \hat{j}$,$\vec{v}_2 = -100 \hat{j}$,and $\vec{v}_3 = V_x \hat{i} + V_y \hat{j}$.By the law of conservation of linear momentum: $\vec{P}_{initial} = \vec{P}_{final}$.$m(100 \hat{i}) = \frac{m}{3}(100 \hat{j}) + \frac{m}{3}(-100 \hat{j}) + \frac{m}{3}(V_x \hat{i} + V_y \hat{j})$.$100 \hat{i} = \frac{1}{3}(V_x \hat{i} + V_y \hat{j})$.Comparing components: $V_x = 300 \, m/s$ and $V_y = 0$.Therefore,the third fragment moves with a velocity of $300 \, m/s$ in the horizontal direction.

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